Find Eigenvectors for x=2 of Matrix A: Help!

[moco89]

Homework Statement

Given the matrix

1 1 1
-1 3 1
-1 1 3

x=3 is an eigenvalue and (1,1,1) is a corresponding eigenvector
x=2 is an eigenvalue of A of multiplicity 2

Find the eigenvector(s) corresponding to x=2

The Attempt at a Solution

(A-AI)=
-1 1 1
-1 1 1
-1 1 1

which row reduces to

-1 1 1
0 0 0
0 0 0

and the formula for the eigenvector is therefore x1=x2+x3

so I get (2,1,1) for the eigenvector but when I solve for the second eigenvector I basically get no solution

-1 1 1 | 2
-1 1 1 | 1
-1 1 1 | 1

R1-R2=>R1
====>

0 0 0 |1
-1 1 1 | 1
-1 1 1 | 1

R2-R3=>R3
====>

0 0 0 | 1
-1 1 1 | 1
0 0 0 | 0

What am I doing wrong? Help please...I have a final tomorrow! Thanks

 

[Mark44]

Homework Statement

Given the matrix

1 1 1
-1 3 1
-1 1 3

x=3 is an eigenvalue and (1,1,1) is a corresponding eigenvector
x=2 is an eigenvalue of A of multiplicity 2

Find the eigenvector(s) corresponding to x=2

The Attempt at a Solution

(A-AI)=
-1 1 1
-1 1 1
-1 1 1

You mean A - 2I.

which row reduces to

-1 1 1
0 0 0
0 0 0

and the formula for the eigenvector is therefore x1=x2+x3

Actually, you should get
x1 = x2 + x3
x2 = x2
x3 = ... + x3

So all vectors in the eigenspace of this eigenvalue (lambda = 2) are linear combinations of <1, 1, 0> and <1, 0, 1>.

so I get (2,1,1) for the eigenvector but when I solve for the second eigenvector I basically get no solution

-1 1 1 | 2
-1 1 1 | 1
-1 1 1 | 1

R1-R2=>R1
====>

0 0 0 |1
-1 1 1 | 1
-1 1 1 | 1

R2-R3=>R3
====>

0 0 0 | 1
-1 1 1 | 1
0 0 0 | 0

What am I doing wrong? Help please...I have a final tomorrow! Thanks

 

[moco89]

You mean A - 2I.

Actually, you should get
x1 = x2 + x3
x2 = x2
x3 = ... + x3

So all vectors in the eigenspace of this eigenvalue (lambda = 2) are linear combinations of <1, 1, 0> and <1, 0, 1>.

Why does x2=x2 and x3=...+x3?

I still don't understand why the eigenvectors are [1,1,0] and [1,0,1] instead of [2,1,1] and [3,1,2] for example.

I am missing a key point here, and I really need your help.

 

[vela]

Mark was just writing down the obviously true statements that x₂=x₂ and x₃=x₃ to suggest what the solution is in terms of vectors.

\vec{x} = \begin{bmatrix}x_1 \\ x_2 \\ x_3 \end{bmatrix} = x_2\begin{bmatrix}1 \\ 1 \\ 0 \end{bmatrix} +<br /> x_3\begin{bmatrix}1 \\ 0 \\ 1 \end{bmatrix}

Note the two vectors [2,1,1] and [3,1,2] are linear combinations of [1,1,0] and [1,0,1]. All four vectors are eigenvectors.

 

[moco89]

Mark was just writing down the obviously true statements that x₂=x₂ and x₃=x₃ to suggest what the solution is in terms of vectors.

\vec{x} = \begin{bmatrix}x_1 \\ x_2 \\ x_3 \end{bmatrix} = x_2\begin{bmatrix}1 \\ 1 \\ 0 \end{bmatrix} +<br /> x_3\begin{bmatrix}1 \\ 0 \\ 1 \end{bmatrix}

Note the two vectors [2,1,1] and [3,1,2] are linear combinations of [1,1,0] and [1,0,1]. All four vectors are eigenvectors.

Thanks a lot to both of you!

You helped me a lot.

 

[HallsofIvy]

Homework Statement

Given the matrix

1 1 1
-1 3 1
-1 1 3

x=3 is an eigenvalue and (1,1,1) is a corresponding eigenvector
x=2 is an eigenvalue of A of multiplicity 2

Find the eigenvector(s) corresponding to x=2

The Attempt at a Solution

(A-AI)=
-1 1 1
-1 1 1
-1 1 1

which row reduces to

-1 1 1
0 0 0
0 0 0

and the formula for the eigenvector is therefore x1=x2+x3

so I get (2,1,1) for the eigenvector but when I solve for the second eigenvector I basically get no solution

(2, 1, 1) is ONE solution to x1= x2+ x3 but not all- nor are all solutions multiples of that one. x1= x2+ x3 means your eigenvectors must be of the form (x1, x2, x3)= (x2+ x3, x2, x3)= (x2, x2, 0)+ (x3, 0, x3)= x2(1, 1, 0)+ x3(1, 0, 1).

-1 1 1 | 2
-1 1 1 | 1
-1 1 1 | 1

R1-R2=>R1
====>

0 0 0 |1
-1 1 1 | 1
-1 1 1 | 1

R2-R3=>R3
====>

0 0 0 | 1
-1 1 1 | 1
0 0 0 | 0

What am I doing wrong? Help please...I have a final tomorrow! Thanks